这个C/ c++代码在Python中的等效代码是什么?
void foo()
{
static int counter = 0;
counter++;
printf("counter is %d\n", counter);
}
具体来说,如何在函数级别实现静态成员,而不是在类级别?将函数放入类中会改变什么吗?
这个C/ c++代码在Python中的等效代码是什么?
void foo()
{
static int counter = 0;
counter++;
printf("counter is %d\n", counter);
}
具体来说,如何在函数级别实现静态成员,而不是在类级别?将函数放入类中会改变什么吗?
当前回答
Python没有静态变量,但你可以通过定义一个可调用的类对象,然后将其用作函数来伪装它。也可以看看这个答案。
class Foo(object):
# Class variable, shared by all instances of this class
counter = 0
def __call__(self):
Foo.counter += 1
print Foo.counter
# Create an object instance of class "Foo," called "foo"
foo = Foo()
# Make calls to the "__call__" method, via the object's name itself
foo() #prints 1
foo() #prints 2
foo() #prints 3
注意,__call__使类(对象)的实例可以通过自己的名称调用。这就是为什么上面调用foo()会调用类的__call__方法。从文档中可以看到:
任意类的实例都可以通过在类中定义__call__()方法来实现可调用。
其他回答
鉴于这个问题,我可以提出另一个可能更好用的替代方案,对方法和函数看起来都是一样的:
@static_var2('seed',0)
def funccounter(statics, add=1):
statics.seed += add
return statics.seed
print funccounter() #1
print funccounter(add=2) #3
print funccounter() #4
class ACircle(object):
@static_var2('seed',0)
def counter(statics, self, add=1):
statics.seed += add
return statics.seed
c = ACircle()
print c.counter() #1
print c.counter(add=2) #3
print c.counter() #4
d = ACircle()
print d.counter() #5
print d.counter(add=2) #7
print d.counter() #8
如果你喜欢这种用法,下面是它的实现:
class StaticMan(object):
def __init__(self):
self.__dict__['_d'] = {}
def __getattr__(self, name):
return self.__dict__['_d'][name]
def __getitem__(self, name):
return self.__dict__['_d'][name]
def __setattr__(self, name, val):
self.__dict__['_d'][name] = val
def __setitem__(self, name, val):
self.__dict__['_d'][name] = val
def static_var2(name, val):
def decorator(original):
if not hasattr(original, ':staticman'):
def wrapped(*args, **kwargs):
return original(getattr(wrapped, ':staticman'), *args, **kwargs)
setattr(wrapped, ':staticman', StaticMan())
f = wrapped
else:
f = original #already wrapped
getattr(f, ':staticman')[name] = val
return f
return decorator
下面是一个完全封装的版本,不需要外部初始化调用:
def fn():
fn.counter=vars(fn).setdefault('counter',-1)
fn.counter+=1
print (fn.counter)
在Python中,函数是对象,我们可以简单地通过特殊属性__dict__向它们添加或修补成员变量。内置的vars()返回特殊属性__dict__。
EDIT:注意,与另一种try不同:除了AttributeError答案外,使用这种方法,变量将始终为初始化后的代码逻辑做好准备。我认为try:except AttributeError替代以下将不那么干和/或有尴尬的流程:
def Fibonacci(n):
if n<2: return n
Fibonacci.memo=vars(Fibonacci).setdefault('memo',{}) # use static variable to hold a results cache
return Fibonacci.memo.setdefault(n,Fibonacci(n-1)+Fibonacci(n-2)) # lookup result in cache, if not available then calculate and store it
EDIT2:当函数将从多个位置调用时,我只推荐上述方法。如果函数只在一个地方被调用,最好使用nonlocal:
def TheOnlyPlaceStaticFunctionIsCalled():
memo={}
def Fibonacci(n):
nonlocal memo # required in Python3. Python2 can see memo
if n<2: return n
return memo.setdefault(n,Fibonacci(n-1)+Fibonacci(n-2))
...
print (Fibonacci(200))
...
你也可以考虑:
def foo():
try:
foo.counter += 1
except AttributeError:
foo.counter = 1
推理:
非常python化(“请求原谅而不是允许”) 使用异常(只抛出一次)而不是if分支(考虑StopIteration异常)
我写了一个简单的函数来使用静态变量:
def Static():
### get the func object by which Static() is called.
from inspect import currentframe, getframeinfo
caller = currentframe().f_back
func_name = getframeinfo(caller)[2]
# print(func_name)
caller = caller.f_back
func = caller.f_locals.get(
func_name, caller.f_globals.get(
func_name
)
)
class StaticVars:
def has(self, varName):
return hasattr(self, varName)
def declare(self, varName, value):
if not self.has(varName):
setattr(self, varName, value)
if hasattr(func, "staticVars"):
return func.staticVars
else:
# add an attribute to func
func.staticVars = StaticVars()
return func.staticVars
使用方法:
def myfunc(arg):
if Static().has('test1'):
Static().test += 1
else:
Static().test = 1
print(Static().test)
# declare() only takes effect in the first time for each static variable.
Static().declare('test2', 1)
print(Static().test2)
Static().test2 += 1
另一个(不推荐!)对https://stackoverflow.com/a/279598/916373这样的可调用对象的扭曲,如果您不介意使用一个时髦的调用签名的话
class foo(object):
counter = 0;
@staticmethod
def __call__():
foo.counter += 1
print "counter is %i" % foo.counter
>>> foo()()
counter is 1
>>> foo()()
counter is 2